Problem: Graph this system of equations and solve. $8x-y = 4$ $x+y = 5$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Answer: Convert the first equation, $8x-y = 4$ , to slope-intercept form. $y = 8 x - 4$ The y-intercept for the first equation is $-4$ , so the first line must pass through the point $(0, -4)$ The slope for the first equation is $8$ . Remember that the slope tells you rise over run. So in this case for every $8$ positions you move up $1$ position to the right. $8$ positions up from $(0, -4)$ is $(1, 4)$ Graph the blue line so it passes through $(0, -4)$ and $(1, 4)$ Convert the second equation, $x+y = 5$ , to slope-intercept form. $y = - x + 5$ The y-intercept for the second equation is $5$ , so the second line must pass through the point $(0, 5)$ The slope for the second equation is $-1$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move down (because it's negative) You must also move $1$ position to the right. $1$ position to the right. $1$ position down from $(0, 5)$ is $(1, 4)$ Graph the green line so it passes through $(0, 5)$ and $(1, 4)$ The solution is the point where the two lines intersect. The lines intersect at $(1, 4)$.